Question: Simplify the following expression and state the conditions under which the simplification is valid. You can assume that $a \neq 0$. $x = \dfrac{a - 1}{-10a + 30} \div \dfrac{a^2 - 1}{10a - 30} $
Answer: Dividing by an expression is the same as multiplying by its inverse. $x = \dfrac{a - 1}{-10a + 30} \times \dfrac{10a - 30}{a^2 - 1} $ First factor the quadratic. $x = \dfrac{a - 1}{-10a + 30} \times \dfrac{10a - 30}{(a - 1)(a + 1)} $ Then factor out any other terms. $x = \dfrac{a - 1}{-10(a - 3)} \times \dfrac{10(a - 3)}{(a - 1)(a + 1)} $ Then multiply the two numerators and multiply the two denominators. $x = \dfrac{ (a - 1) \times 10(a - 3) } { -10(a - 3) \times (a - 1)(a + 1) } $ $x = \dfrac{ 10(a - 1)(a - 3)}{ -10(a - 3)(a - 1)(a + 1)} $ Notice that $(a - 3)$ and $(a - 1)$ appear in both the numerator and denominator so we can cancel them. $x = \dfrac{ 10\cancel{(a - 1)}(a - 3)}{ -10(a - 3)\cancel{(a - 1)}(a + 1)} $ We are dividing by $a - 1$ , so $a - 1 \neq 0$ Therefore, $a \neq 1$ $x = \dfrac{ 10\cancel{(a - 1)}\cancel{(a - 3)}}{ -10\cancel{(a - 3)}\cancel{(a - 1)}(a + 1)} $ We are dividing by $a - 3$ , so $a - 3 \neq 0$ Therefore, $a \neq 3$ $x = \dfrac{10}{-10(a + 1)} $ $x = \dfrac{-1}{a + 1} ; \space a \neq 1 ; \space a \neq 3 $